Tính :23+3/4-(1/2:3/4)
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\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
Số khá xấu. Bạn coi lại đề xem có viết nhầm biểu thức không?
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{23.24.25}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{23.24.25}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{23.24}-\frac{1}{24.25}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}-\frac{1}{2.3}\right)-\left(\frac{1}{3.4}-\frac{1}{3.4}\right)-\left(\frac{1}{4.5}-\frac{1}{4.5}\right)-....-\left(\frac{1}{23.24}-\frac{1}{23.24}\right)-\frac{1}{24.25}\)
\(=\frac{1}{1.2}-\frac{1}{24.25}\)
\(=\frac{1}{2}-\frac{1}{600}=\frac{299}{600}\)
Vậy : \(A=\frac{299}{600}\)
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
A = 2/1*5 + 2/5*9 + ... + 2/101*105
= 1/2(4/1*5 + 4/5*9 + ... + 4/101*105)
= 1/2(1 - 1/5 + 1/5 - 1/9 + ... + 1/101 - 1/105)
= 1/2(1 - 1/105)
= 1/2 * 104/105 = 52/105
Sửa câu b. Phân số thứ 2 phải là 4/5*8
B = 4/2*5 + 4/5*8 + ... + 4/47*50
= 4/3(3/2*5 + 3/5*8 + ... + 3/47*50)
= 4/3(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/47 - 1/50)
= 4/3(1/2 - 1/50)
= 4/3 * 24/50 = 16/25
5:
a: \(3^{2n}=\left(3^2\right)^n=9^n\)
\(\left(2^{3n}\right)=\left(2^3\right)^n=8^n\)
=>\(3^{2n}>2^{3n}\)
b: \(199^{20}=\left(199^4\right)^5=1568239201^5\)
\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)
mà \(1568239201< 8036054027\)
nên \(199^{20}< 2003^{15}\)
4: \(100< 5^{2x-1}< 5^6\)
mà \(25< 100< 125\)
nên \(125< 5^{2x-1}< 5^6\)
=>3<2x-1<6
=>4<2x<7
=>2<x<7/2
mà x nguyên
nên x=3
c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
C = 2-3 + (52)3.5-3 + 4-3.16 - 2.32 - 105.(\(\dfrac{24}{51}\))0
C = \(\dfrac{1}{8}\) + 56.5-3 + 4-3.42 - 2.9 - 105.1
C = \(\dfrac{1}{8}\) + 53 + \(\dfrac{1}{4}\) - 18 - 105
C = (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\)) - (105 - 125 + 18)
C = \(\dfrac{3}{8}\) - (-20 + 18)
C = \(\dfrac{3}{8}\) + 2
C = \(\dfrac{19}{8}\)
\(2^3+\frac{3}{4}-\left(\frac{1}{2}:\frac{3}{4}\right)=8+\frac{3}{4}-\left(\frac{1}{2}:\frac{3}{4}\right)=8+\frac{3}{4}-\frac{2}{3}=8+\frac{1}{12}=\frac{97}{12}\)
học tốt
\(2^3+\frac{3}{4}-\left(\frac{1}{2}:\frac{3}{4}\right)\)
\(=8+\frac{3}{4}-\left(\frac{1}{2}.\frac{4}{3}\right)\)
\(=8+\frac{3}{4}-\frac{2}{3}\)
\(=\frac{96}{12}+\frac{9}{12}-\frac{8}{12}=\frac{97}{12}\)